We use the isomorphism \(W\cong V\rtimes S_n\) , where \(V=\mathbb {F}_2^{n-1}\) . The action of \(\sigma \in S_n\)  on \(V=\langle v_1,\ldots v_{n-1}\rangle \)  is by
\(v_i\sigma ={\left\lbrace \begin{array}{ll}v_{i\sigma },&\text{ if }n\sigma =n,\\v_{n\sigma },&\text{ if }i\sigma =n,\text{ while }\\v_{i\sigma }+v_{n\sigma },&\text{ if }i\sigma ,\,n\sigma <n.\end{array}\right.}\) 
The claim about the Gelfand property of \(P\)  is covered by Theorem REF . From now on we assume \(n/2<k<n-1\) . Then \(P\)  is the Weyl group of the decomposable root system of type \(A_{k-1}\oplus D_{n-k}\)  (with minor notational changes due to degeneration if \(k=n-3,n-2\) ) and \(P\cong S_{k}\times (V_0\rtimes S_{n-k})\) , where \(V_0=\langle v_{k+1},\ldots v_{n-1}\rangle \) . Let \(Q=N_W(P)\) . As \(k>n/2\) , \(Q= N_V(P)P=\langle v_1+v_2+\cdots +v_k\rangle P\) . So let \(V_1=Q\cap V=\langle v_1+v_2+\cdots +v_k, v_{k+1},\ldots ,v_{n-1}\rangle \) .
To determine \(1_{Q}^W\)  we first decompose \(1_{Q}^{VQ}\) . Using a special case of Mackey's theorem [1]}, \({1_{Q}^{VQ}}\downarrow _V=1_{V_1}^V\) , whose constituents correspond to \(E\subseteq \lbrace 1,\ldots , k\rbrace \)  of cardinality \(|E|\equiv k\pmod {2}\) . Namely, for each such \(E\)  let \(\eta _E\in \operatorname{\rm Irr}(V)\)  be such that \(v_j\in \operatorname{\rm Ker}(\eta _E)\)  if and only if \(j\in E\)  or \(j>k\) . Then \(1_{V_1}^V=\sum _E\eta _E\) . Note that the orbit of a subset \(E\subseteq \lbrace 1,\ldots ,k\rbrace \)  under the action of \(S_k\times S_{n-k}\)  consists of the subsets of the same cardinality. So \(1_{Q}^{VQ}=\sum _{i=0}^{[k/2]}\chi _i\) , where \({\chi _i}_V=\sum _{|E|=k-2i}\eta _E\) . Fix \(E=\lbrace 2i+1,\ldots ,k\rbrace \)  and the inertia subgroup of \(\eta _E\)  in \(VQ\)  is \(M=V(S_{2i}\times S_{k-2i}\times S_{n-k})=I_{VQ}(\eta _E)\) . If \(\nu _E\in \operatorname{\rm Irr}(M)\)  denotes the extension of \(\eta _E\)  to \(M\)  which is trivial on \(S_{2i}\times S_{k-2i}\times S_{n-k}\) , then \(\chi _i=\nu _E^{VQ}\)  is irreducible by[1]}.
We claim that \(\chi _i^W\)  are all multiplicity free and \((\chi _i^W,\chi _j^W)>0\)  if and only if \(i=j\)  or \(2i+2j=n\) . Justifying this claim is enough, as this latter possibility can occur only for even \(n\) . But if \(n\)  is divisible by 4 then \(k=n/2+1\)  odd and the largest distinct \(2i,2j\)  are \(k-1,k-3\) , their sum is \(2k-4=n-2\) , so even this is impossible. (But for other even \(n\)  such pair of distinct even numbers do occur. If \(k\)  is even, \(k>n/2\)  then \(i=k/2>(n-k)/2=j\) , and if \(k>2[n/4]+1\)  is odd then \(i=(k-1)/2>(n-k+1)/2=j\) .)
To prove multiplicity freeness, let \(i\le k/2\)  be fixed, \(E=\lbrace 2i+1,\ldots ,k\rbrace \)  and \(L=S_{2i}\times VS_{n-2i}=I_W(\eta _E)\) . Then \(\eta _E\)  extends to \(\mu _E\in \operatorname{\rm Irr}(L)\)  such that \(S_{2i}\times S_{n-2i}\subseteq \operatorname{\rm Ker}(\mu _E)\) . Of course, \(\mu _E\)  also extends \(\nu _E\) . The other constituents of \(\eta _E^L\)  are \(\mu _E\varphi \) , such that \(\varphi \in \operatorname{\rm Irr}(L)\)  and \(Ker(\varphi )\supseteq V\) , in particular, all the irreducible constituents of \(\nu _E^L\)  are of this form. By [1]}, each \((\mu _E\varphi )^W\)  is irreducible. Now, \(\chi _i^W=(\nu _E^{VQ})^W=\nu _E^W=(\nu _E^L)^W\)  and, by Frobenius reciprocity,
\((\nu _E^L,\mu _E\varphi )=(\nu _E,(\mu _E\varphi )_M)=(\nu _E,\nu _E\varphi _M)=(1_M,\varphi _M)=(1_M^L,\varphi )\le 1,\) 
because \(M\le L\)  is a Gelfand subgroup. (Consider \(M/VS_{2i}\cong S_{k-2i}\times S_{n-k}\le S_{n-2i}\cong L/VS_{2i}\) .) That is, \(\nu _E^L\)  is multiplicity free. As each of its irreducible constituents are of form \(\mu _E\varphi \)  and these induce irreducibly to \(W\) , so we also get that \(\chi _i^W=\nu _E^W\)  is multiplicity free, as required.
Given \(i\ne j\)  the characters \(\chi _i^W\)  and \(\chi _j^W\)  share no common constituent if the underlying \(\eta _{E_i}\)  and \(\eta _{E_j}\)  are not \(W\) -conjugate for \(E_i=\lbrace 2i+1,\ldots ,k\rbrace ,\,E_j=\lbrace 2j+1,\ldots ,k\rbrace \) . Suppose \(\sigma \)  is such that \(\eta _{E_i}^\sigma =\eta _{E_j}\) . If \(\sigma n=n\)  then
\(\lbrace \sigma (1),\ldots ,\sigma (2i)\rbrace =\lbrace f\mid v_f\notin \operatorname{\rm Ker}(\eta _{E_i}^\sigma ) \rbrace =\lbrace f\mid v_f\notin \operatorname{\rm Ker}(\eta _{E_j}) \rbrace =\lbrace 1,\ldots ,2j \rbrace \) 
have the same cardinality, so \(i=j\) . If \(\sigma m=n\) , \(2i<m<n\)  then \(v_{\sigma n}=\sigma v_m\in \operatorname{\rm Ker}(\eta _{E_i}^\sigma )\) , so
\(\lbrace \sigma (1),\ldots ,\sigma (2i)\rbrace =\lbrace f\mid v_f\notin \operatorname{\rm Ker}(\eta _{E_i}^\sigma ) \rbrace =\lbrace f\mid v_f\notin \operatorname{\rm Ker}(\eta _{E_j}) \rbrace =\lbrace 1,\ldots ,2j \rbrace ,\) 
again and \(i=j\) . Finally, if \(\sigma m=n\) , \(m\le 2i\)  then \(v_{\sigma n}=\sigma v_m\notin \operatorname{\rm Ker}(\eta _{E_i}^\sigma )\) , so
\(\lbrace \sigma (n),\sigma (2i+1),\ldots \sigma (n-1)\rbrace =\lbrace f\mid v_f\notin \operatorname{\rm Ker}(\eta _{E_i}^\sigma ) \rbrace =\lbrace f\mid v_f\notin \operatorname{\rm Ker}(\eta _{E_j}) \rbrace =\lbrace 1,\ldots ,2j \rbrace \) 
have the same cardinality \(n-2i=2j\) , that is \(n=2i+2j\) , as required. The claim is established.
Let now \(n\)  be even, \(j=[k/2]>i=n/2-j>0\)  and \(E_i=\lbrace 2i+1,\ldots ,k\rbrace ,\,E_j=\lbrace 2j+1,\ldots ,k\rbrace \) . Let \(\sigma \)  be the transposition \((1,2j+1)(2,2j+2)\cdots (2i,n)\) , hence \(\eta _{E_j}^\sigma =\eta _{E_i}\) . If \(M_i=I_{VQ}(\eta _{E_i})\)  and \(M_j=I_{VQ}(\eta _{E_j})\)  then \(\nu _{E_i}\downarrow _{M_j^\sigma \cap M}=\nu _{E_j}^\sigma \downarrow _{M_j^\sigma \cap M}\) . By using Mackey's theorem[1]}, we get
\((\chi _i^W,\chi _j^W)=(\nu _{E_i}^W,\nu _{E_j}^W)=(\nu _{E_i},\nu _{E_j}^W\downarrow _{M_i})=\sum _{W=\cup M_j g M_i}(\nu _{E_i},(\nu _{E_j}^g\downarrow _{M_j^g\cap M_i})^{M_i}).\) 
Among the summands is
\((\nu _{E_i},(\nu _{E_j}^\sigma \downarrow _{M_j^\sigma \cap M_i})^{M_i})=(\nu _{E_i},(\nu _{E_i}\downarrow _{M_j^\sigma \cap M_i})^{M_i})=(\nu _{E_i}\downarrow _{M_j^\sigma \cap {M_i}},\nu _{E_i}\downarrow _{M_j^\sigma \cap {M_i}})=1,\) 
hence \(1_{Q}^W=\sum _{i=0}^{[k/2]} \chi _i^W\)  is not multiplicity free.
